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Appendix VL 1085 Referring to Fig. 959. Let a load of w Ibs. per foot run be advanced from the left up to any point on the beam. It is required to find the shearing force at the section K. Diagram (i). The load is advanced up to K. Reaction at A has the general form <wx (I - - j ~- / for every case. Reaction at B, which is shear at K is wx. - -r /. The shear at K is less than Diagram (2). The load is short of ic. before, because both reactions are less. Diagram (3). When loaded beyond K. Both reactions are in- creased ; but the shear at K is actually less than in (i) for all the load beyond K has to be deducted in arriving at the shearing force, whereas only part of it adds to the reaction. Therefore the greatest shear at K occurs when the load travels exactly up to that point. Next take various positions for K between A and B. The ordinates of the curve of maximum shear for these points will be, from dia- gram (2) . , wx x - -~ / = 2 r- 2/ But x is the abscissa from A .*. Ordinates cc x* and the curve is drawn at (2). Pp. 446, 464, 4jo, and 885. Link Polygons.It is re- quired to add, that is find, the resultant of a number of forces inclined at various angles, but whose directions are co-planar, that is, lie within the same plane. The forces may meet in one point, and the method on p. 464 be found convenient; but if the directions are varied in any possible co-planar manner, the construction known as the link or funicular polygon must be adopted. Let the lines ABODE, Fig. 960, represent forces in direction, sense, and magnitude whose resultant is required. Draw the unclosed polygon o i 2 3 4 5 by lines parallel to the forces, whose lengths are also proportional to their magnitudes. The closing line R will naturally shew the magnitude and direction of the resultant, and the sense will be reversed, as shewn by the arrow. We must next shew on the space diagram the exact position of R. Choose any pole O in the vector figure, either in or out of the